从R中的日期中删除年份 [英] Remove year from dates in R

问题描述

我有20年的气象数据，但我只对每年的模式感兴趣。例如，我不在乎1995年6月与2011年6月有何不同。相反，我想在6月1日使用20个值，在6月2日使用20个值，等等。

I have 20 years worth of weather data, but I'm only interested in the patterns per year. I don't care how June 1995 is different from June 2011, for example. Instead, I want to have 20 values for June 1, 20 values for June 2, etc.

我的问题：如何删除日期对象的年份部分，保留月份和日期，同时还保留日期的顺序属性？我的最终目标是一长串重复的mm / dd值，每个值对应于结果变量。我将把mm / dd视为因子，但顺序正确。

My question: How do I drop the year portion of a date object, keep the month AND day, while also maintaining the sequential properties of dates? My ultimate goal is a long list of repeated mm/dd values corresponding each to the outcome variable. I'll treat the mm/dd like factors, but in the correct order.

# Given this:
as.Date(c("2014-06-01","1993-06-01", "2013-06-03", "1999-01-31"), "%Y-%m-%d")
# I want to get this:
"06-01" "06-01" "06-03" "01-31"
# That will sort like this
"01-31" "06-01" "06-01" "06-03"

小技巧使用sub（）删除年份并将破折号转换为十进制不起作用，因为这时该月的1号与该月的10号相同。我还尝试过将日期转换成字符串，删除年份，然后将其转换回日期……这一切都使2014年成为现实。

Little hacks like using sub() to drop the year and convert the dash to a decimal doesn't work because then the 1st of the month is the same as the 10th of the month. I also tried turning the dates into character strings, removing the year, and then turning it back into a date... that just made everything year 2014.

推荐答案

这项工作吗？

temp<-as.Date(c("2014-06-01","1993-06-01", "2013-06-03", "1999-01-31"), "%Y-%m-%d")

x<-format(temp, format="%m-%d")

 x
[1] "06-01" "06-01" "06-03" "01-31"


sort(x)
[1] "01-31" "06-01" "06-01" "06-03"

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